Can you write a balanced equation for BCl3 (g)+H2O (l)-------> H3BO3 (s)+HCl (g)?
The balanced equation is BCl₃(g) + 3H₂O(l) →H₃BO₃(s) + 3HCl(aq)
The usual procedure is to balance all atoms other than H and O; then balance O; then balance H.
Start with most complicated-looking molecule in the equation. This looks like H₃BO₃.
1. Balance all atoms other than H and O.
Put a 1 in front of the H₃BO₃. We start with
BCl₃ + H₂O → 1 H₃BO₃ + HCl
Since we have fixed one B atom on the right, we need one B atom on the left. We put a 1 in front of the BCl₃.
1 BCl₃ + H₂O → 1 H₃BO₃ + HCl
2. Balance O.
We have fixed 3 O atoms on the right, so we need 3 O atoms on the left. We put a 3 in front of the H₂O.
1 BCl₃ + 3 H₂O → 1 H₃BO₃ + HCl
3. Balance H.
Now we have fixed 6 H atoms on the left, so we need 6 H atoms on the right. We already have 3 H atoms on the H₃BO₃. We need 3 more H atoms, so we place a 3 in front of the HCl.
1 BCl₃ + 3H₂O → 1 H₃BO₃ + 3 HCl
4. Check that Equation is Balanced.
Since every formula has a coefficient, we should now have a balanced equation. Let’s check:
On the left: 1 B; 3 Cl; 6 H; 3 O.
On the right: 6 H; 1 B; 3 O; 3 Cl.
Our balanced equation is
BCl₃(g) + 3H₂O(l) → H₃BO₃(s) + HCl(aq)
Note that the HCl is so reactive and so soluble in water that it will be present as HCl(aq), not as HCl(g).
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